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chubhunter [2.5K]
4 years ago
11

The graph of a secant function shows no functional values on (−7, 7) . The asymptotes of the function occur every π units. Which

function represents the secant function described?
f(x)=sec(x/2)+7



f(x)=7sec(x/2)



f(x)=7sec(x)



f(x)=sec(x)+7

Mathematics
2 answers:
Nadya [2.5K]4 years ago
6 0

Answer:

f(x) = 7 sec(x)

Step-by-step explanation:

The range of the function is less than -7 or greater than 7.  So the midline is y=0 and the amplitude is 7.

There are two asymptotes per period, so the period of the function is 2π.

Therefore, the function is:

f(x) = A sec(2π / T x)

f(x) = 7 sec(2π / (2π) x)

f(x) = 7 sec(x)

ArbitrLikvidat [17]4 years ago
4 0

Answer:

  f(x) = 7sec(x)

Step-by-step explanation:

The parent secant function has an excluded range of (-1, 1), so the one we're seeking has been <em>scaled vertically</em> by a factor of 7. It will look like ...

  f(x) = 7sec( )

__

The parent secant function has vertical asymptotes at odd multiples of π/2, so are separated by π units. Thus, there is <em>no horizontal scaling</em>.

The function described is f(x) = 7sec(x).

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